Let $a(x)=2x^7-3x^5+5x^3-7x^2$, and $b(x)=x^7+x^5+x^3+x^2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{2x^7-3x^5+5x^3-7x^2}{x^7+x^5+x^3+x^2}$ : We divide ${x^7}$ into ${2x^7}$ to get ${2}$ : $ \hphantom{1567|144447777784} {2}\\ {{{x^7}+x^5+x^3+x^2}}|\overline{{2x^7}-3x^5+5x^3-7x^2}\\ \hphantom{37....899888........|}\llap{-}\underline{(2x^7+2x^5+2x^3+2x^2)}\\ \hphantom{37|3....998888888........}{-5x^5+3x^3 -9x^2}\\ $ [What did we do here?] The process stops here because $x^7+x^5+x^3+x^2$ is a polynomial of the seventh degree and $-5x^5+3x^3 -9x^2$ is a polynomial of the fifth degree. So it follows that ${r(x)}={-5x^5+3x^3 -9x^2}$, ${q(x)}={2}$, and $ \dfrac{2x^7-3x^5+5x^3-7x^2}{x^7+x^5+x^3+x^2}={2}+\dfrac{{-5x^5+3x^3 -9x^2}}{x^7+x^5+x^3+x^2}$ To conclude, $q(x)=2$ $r(x)=-5x^5+3x^3 -9x^2$